Domino odds
January 23rd, 2010 at 9:45:08 PM
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I was wondering if someone can help me with an odds problem.
I'm trying to find out how to figure out odds on a set of dominoes.
For example, using a double nine 55 tile set and each player pulls x no. of tiles, what are the odds of pulling 4 tiles with the same no. or pulling multiple double tiles.
Thanks.
I'm trying to find out how to figure out odds on a set of dominoes.
For example, using a double nine 55 tile set and each player pulls x no. of tiles, what are the odds of pulling 4 tiles with the same no. or pulling multiple double tiles.
Thanks.
January 24th, 2010 at 1:21:22 PM
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I have the second part of the problem solved, it's a combinatorics formula.
Out of 55 double-nine dominoes, 10 are doubles, which includes double blank.
When drawing one domino, the probability of picking a double is 10/55, or 18.18%.
When drawing two dominoes, the probability of picking exactly one double is
which is 30.30%. That is, picking one of 45 non-doubles, and picking one of 10 doubles, divided into picking 2 of the 55 domino pool. You can determine the probability of drawing at least one double by taking 1 minus the probability of drawing no doubles:
which is 1-(0.666) = 0.333 = 33.33%
To determine the probability of drawing exactly "d" doubles out of "p" dominoes drawn, simply use this formula:
p = dominoes to pick, d = doubles to pick, n = non-doubles to pick, when d+n = p
Out of 55 double-nine dominoes, 10 are doubles, which includes double blank.
When drawing one domino, the probability of picking a double is 10/55, or 18.18%.
When drawing two dominoes, the probability of picking exactly one double is
(45C1 x 10C1)
-----------------
55C2
which is 30.30%. That is, picking one of 45 non-doubles, and picking one of 10 doubles, divided into picking 2 of the 55 domino pool. You can determine the probability of drawing at least one double by taking 1 minus the probability of drawing no doubles:
45C2
1 - -----------------
55C2
which is 1-(0.666) = 0.333 = 33.33%
To determine the probability of drawing exactly "d" doubles out of "p" dominoes drawn, simply use this formula:
p = dominoes to pick, d = doubles to pick, n = non-doubles to pick, when d+n = p
(45Cn x 10Cd)
-----------------
55Cp
-Dween!
January 24th, 2010 at 3:51:23 PM
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Thanks that is a big help. I came up with a .00007244 probability of getting 5 doubles in a 5 tile hand and .0027 for a 4 double hand. I am assuming the formula doesn't work for choosing individual #'s. That seems to be a little more complicated due to the fact that dominoes have two values.
January 25th, 2010 at 6:33:04 AM
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As for figuring out drawing x dominoes and having at least y of them with at least one matching number, I ran a VBA simulation for that. Before giving the results, I have a simple case that can be figured out mathematically.
In drawing 2 dominoes, what are the chances that they will share a number?
45/55 will be a 2-number domino, which means 18/54 dominoes will have one number in common with the original.
10/55 will be a double, that is, only have a single number. 9/54 dominoes will have that number in common with the original.Adding those numbers gives the probability of the above question, 0.303030, or 30.3%
A simulation in VBA over 1,000,000 iterations gives 302,915 successful matches, which corresponds to the percentage given in the mathematical formula above.
As for more matches over more draws, VBA will have to take care of that. I have run each case over 1,000,000 iterations. When drawing x dominoes, what are the chances you will have at least y dominoes with a number in common? Here are the results:
The probability of matching a number on at least 7 dominoes, when drawing 7, 8, 9, or 10 dominoes, is less than 0.1%.
To determine the probability that none have a number in common, subtract the Match 2+ value from 100%.
Because of the double dominoes, and the fact that we are looking for any common number matches, figuring out a formula is a bit difficult.
I hope this information is helpful... and I hope my simulation is correct! Any data confirmation would be much appreciated.
In drawing 2 dominoes, what are the chances that they will share a number?
45/55 will be a 2-number domino, which means 18/54 dominoes will have one number in common with the original.
10/55 will be a double, that is, only have a single number. 9/54 dominoes will have that number in common with the original.
45 18 810 10 9 90
-- x -- = ---- = 0.272727 -- x -- = ---- = 0.030303
55 54 2970 55 54 2970
A simulation in VBA over 1,000,000 iterations gives 302,915 successful matches, which corresponds to the percentage given in the mathematical formula above.
As for more matches over more draws, VBA will have to take care of that. I have run each case over 1,000,000 iterations. When drawing x dominoes, what are the chances you will have at least y dominoes with a number in common? Here are the results:
| Draw 2 | Draw 3 | Draw 4 | Draw 5 | Draw 6 | Draw 7 | Draw 8 | Draw 9 | Draw 10 | |
|---|---|---|---|---|---|---|---|---|---|
| No Match | 69.7% | 31.7% | 8.7% | 1.3% | 0.1% | <0.1% | <0.1% | <0.1% | <0.1% |
| Match 2+ | 30.3% | 68.3% | 91.3% | 98.7% | 99.9% | >99.9% | >99.9% | >99.9% | >99.9% |
| Match 3+ | -- | 4.6% | 16.5% | 35.2% | 57.3% | 76.9% | 90.3% | 97.0% | 99.4% |
| Match 4+ | -- | -- | 0.6% | 2.8% | 7.6% | 15.7% | 27.8% | 42.7% | 58.8% |
| Match 5+ | -- | -- | -- | 0.1% | 0.4% | 1.3% | 3.1% | 6.4% | 11.6% |
| Match 6+ | -- | -- | -- | -- | <0.1% | <0.1% | 0.2% | 0.5% | 1.1% |
The probability of matching a number on at least 7 dominoes, when drawing 7, 8, 9, or 10 dominoes, is less than 0.1%.
To determine the probability that none have a number in common, subtract the Match 2+ value from 100%.
Because of the double dominoes, and the fact that we are looking for any common number matches, figuring out a formula is a bit difficult.
I hope this information is helpful... and I hope my simulation is correct! Any data confirmation would be much appreciated.
-Dween!
